Answer: 19
Step-by-step explanation:
3^3=27, 27-7=20, 20·5=100, 100/10=10.
1+3=4, 4+6=10, 10+9=19, 19-10=9.
10+9=19.
What is the question ill try to help
Answer:
Step-by-step explanation:
Given:
u = 1, 0, -4
In unit vector notation,
u = i + 0j - 4k
Now, to get all unit vectors that are orthogonal to vector u, remember that two vectors are orthogonal if their dot product is zero.
If v = v₁ i + v₂ j + v₃ k is one of those vectors that are orthogonal to u, then
u. v = 0 [<em>substitute for the values of u and v</em>]
=> (i + 0j - 4k) . (v₁ i + v₂ j + v₃ k) = 0 [<em>simplify</em>]
=> v₁ + 0 - 4v₃ = 0
=> v₁ = 4v₃
Plug in the value of v₁ = 4v₃ into vector v as follows
v = 4v₃ i + v₂ j + v₃ k -------------(i)
Equation (i) is the generalized form of all vectors that will be orthogonal to vector u
Now,
Get the generalized unit vector by dividing the equation (i) by the magnitude of the generalized vector form. i.e
Where;
|v| = 
|v| = 
= 
This is the general form of all unit vectors that are orthogonal to vector u
where v₂ and v₃ are non-zero arbitrary real numbers.
Answer: 14/15
This is because to get to 14/15 you must divide by 6 but when getting to 42/45 you must divide 2. Therefore 14/15 is your answer because the problem went to the most reduced fraction.
Answer:
a) W₁ = 78400 [J]
b)Wt = 82320 [J]
Step-by-step explanation:
a) W = ∫ f*dl general expression for work
If we have a chain with density of 10 Kg/m, distributed weight would be
9.8 m/s² * 10 kg = mg
Total length of th chain is 40 m, and the function of y at any time is
f(y) = (40 - y ) mg where ( 40 - y ) is te length of chain to be winded
At the beggining we have to wind 40 meters y = 0 at the end of the proccess y = 40 and there is nothing to wind then:
f(y) = mg* (40 - y )
W₁ = ∫f(y) * dy ⇒ W₁ = ∫₀⁴⁰ mg* (40 - y ) dy ⇒ W₁ = mg [ ∫₀⁴⁰ 40dy - ∫₀⁴⁰ ydy
W₁ = mg [ 40*y |₀⁴⁰ - 1/2 * y² |₀⁴⁰ ⇒ W₁ = mg* [ 40*40 - 1/2 (40)² ]
W₁ = mg * [1/2] W₁ = 10*9,8* ( 800 )
W₁ = 78400 [J]
b) Now we can calculate work to do if we have a 25 block and the chain is weightless
W₂ = ∫ mg* dy ⇒ W₂ = ∫₀⁴⁰ mg*dy ⇒ W₂ = mg y |₀⁴⁰
W₂ = mg* 40 = 10*9.8* 40
W₂ = 3920 [J]
Total work
Wt = W₁ + W₂ ⇒ Wt = 78400 + 3920
Wt = 82320 [J]
